Let \(X\) denote the random variable for scores on the exam. \(X\) is normally distributed with mean 57 and standard deviation 9. Let \(Z\) denote the random variable for scores following the standard normal distribution.
You have
\(\mathbb P(X<66)=\mathbb P\left(\dfrac{X-57}9<\dfrac{66-57}9\right)=\mathbb P(Z<1)\)
You could use a table of z-scores to get a precise answer, or you could apply the empirical rule. You know 100% of the scores are contained within the distribution. The empirical rule states that approximately 68% of them will fall within one standard deviation of the mean, which means 32% fall without, with 16% lying to either side of this range.
\(\underbrace{100\%}_{\text{total}}=\underbrace{16\%}_{Z<-1}+\underbrace{68\%}_{-11}\)
\(\mathbb P(Z<1)\) corresponds to the first two ranges, i.e.
\(\mathbb P(Z<1)=\mathbb P(Z<-1)+\mathbb P(-1
So out of 4000 students, you can expect 84% of them, or about 3360, to score lower than a 66. This means the answer is 3366.
More precisely, \(\mathbb P(Z<1)\approx0.8413=84.13\%\), which means closer to 3366 students would fall in this range.
